110. Balanced Binary Tree - Easy

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

Return false.

 

先求出子树的高度，如果左右子树的高度差超过1，返回false；如果不超过1，向下递归

time: O(nlogn)　　-- 求高度O(logn), 一共n个节点, space: O(logn)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root == null) {
            return true;
        }
        if(Math.abs(depth(root.left) - depth(root.right)) > 1) {
            return false;
        }
        return isBalanced(root.left) && isBalanced(root.right);
    }
    
    public int depth(TreeNode root) {
        if(root == null) {
            return 0;
        }
        return 1 + Math.max(depth(root.left), depth(root.right));
    }
}

 

二刷：optimal

class Solution {
    public boolean isBalanced(TreeNode root) {
        if(root == null) {
            return true;
        }
        return getHeight(root) != -1;
    }
    
    public int getHeight(TreeNode root) {
        if(root == null) {
            return 0;
        }
        int left = getHeight(root.left);
        int right = getHeight(root.right);
        if(left == -1 || right == -1 || Math.abs(left - right) > 1) {
            return -1;
        }
        return 1 + Math.max(left, right);
    }
}